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umka2103 [35]
3 years ago
12

Suppose the foreman had released the box from rest at a height of 0.25 m above the ground. What would the crate's speed be when

it reaches the bottom of the ramp
Physics
1 answer:
Arturiano [62]3 years ago
4 0

Answer:

v = 2.21 m/s

Explanation:

The foreman had released the box from rest at a height of 0.25 m above the ground.

We need to find the speed of the crate when it reaches the bottom of the ramp. Let v is the velocity at the bottom of the ramp. It can be calculated using conservation of energy as follows :

mgh=\dfrac{1}{2}mv^2\\\\v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 0.25} \\\\v=2.21\ m/s

So, its velocity at the bottom of the ramp is 2.21 m/s.

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A train is pulling four train cars and each car has a mass of 40,000 kg. The train is accelerating at 1.1 m/s^2. What is the for
IgorLugansk [536]

Answer:

176,000 N

Explanation:

Newton's second law:

∑F = ma

F = (4 × 40,000 kg) (1.1 m/s²)

F = 176,000 N

8 0
3 years ago
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A particle moves along the curve below. y = sqrt(1 + x^3) As it reaches the point (2, 3), the y-coordinate is increasing at a ra
blagie [28]

Answer:7 cm/s

Explanation:

Given

Particle move along curve

y=\sqrt{1+x^3}

As it reaches the (2,3) its y coordinate is increasing at 14 cm/s

Differentiating y w.r.t time

\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{3x^2}{2\sqrt{1+x^3}}\times \frac{\mathrm{d} x}{\mathrm{d} t}

Now at (2,3)

\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{3\cdot 2^2}{2\sqrt{1+2^3}}\times \frac{\mathrm{d} x}{\mathrm{d} t}

14=\frac{3\times 4}{2\times \sqrt{9}}\times \frac{\mathrm{d} x}{\mathrm{d} t}

\frac{\mathrm{d} x}{\mathrm{d} t}=7 cm/s

7 0
3 years ago
The process by which people walk A) requires only a little energy. B) is powered by the body’s surplus heat energy. C) relies on
Andreyy89
I think the answer would be c
4 0
3 years ago
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Rahul has 2 bulbs connected across two cells in a simple circuits shown. How can he make the bulbs glow dimmer?
Ray Of Light [21]

Answer:

in the parallel connection the light bulbs shine less than in the series connection

Explanation:

In a series circuit the current through the whole circuit is the same, therefore the power (brightness) of each bulb is

           P = i² R

where R is the resistance of each bulb and i the current of the circuit.

If we connect the light bulbs and the cells in parallel, the current in the circuit is the sum of the east that passes through each light bulb,

            i = i₁ + i₂

if the two light bulbs are the same

           i = 2 i₁

           i₁ = i / 2

so the power of each bulb is is

           P = i₁² R

           P = R i² / 4

           P = ¼ P_initial

Therefore we see that in the parallel connection the light bulbs shine less than in the series connection

3 0
3 years ago
Electronics and inhabitants of the International Space Station generate a significant amount of thermal energy that the station
katrin2010 [14]

Answer:

4462.0927 W

Explanation:

\epsilon = Emissivity of the panel = 1

\sigma = Stefan-Boltzmann constant = 5.67\times 10^{-8}\ W/m^2K^4

T = Temperature = (273.15+6)

Area of the panel is given by

A=2\times 1.8\times 3.6\\\Rightarrow A=12.96\ m^2

The power radiated is given by

P=\epsilon \sigma AT^4\\\Rightarrow P=1\times 5.67\times 10^{-8}\times 12.96\times (273.15+6)^4\\\Rightarrow P=4462.0927\ W

The power radiated from each panel is 4462.0927 W

5 0
3 years ago
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